Week 6 Homework
Score: \( 27.2 / 30 \quad 29 / 30 \) answered
Question 18
区0/1 pt \( 5 \rightleftarrows 95 C \)
Constructing a Confidence Interval for a Population Mean
You intend to estimate a population mean \( \mu \) with the following sample.
\begin{tabular}{|l|l|l|l|}
\hline 14 & 12 & 18 & 14 \\
\hline 14 & 16 & 17 & 11 \\
\hline 12 & 15 & 12 & 12 \\
\hline
\end{tabular}
You believe the population is normally distributed. Find the \( 90 \% \) confidence interval for the population mean. Enter your answer as an open-interval (i.e., parentheses) accurate to two decimal places.
\[
90 \% \text { c.1. }=
\]
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Answer
\( \text{CI }= (\overline{x} - t^* \frac{s}{\sqrt{n}}, \overline{x} + t^* \frac{s}{\sqrt{n}}) = (14 - 1.753(0.6175), 14 + 1.753(0.6175)) = (13.08, 14.92) \)
Step 1 ：\( n = 16 \)
Step 2 ：\( \overline{x} = \frac{1}{16}(14 + 12 + 18 + 14 + 14 + 16 + 17 + 11 + 12 + 15 + 12 + 12) = 14 \)
Step 3 ：\( s = \sqrt{\frac{1}{15}((14-14)^2 + (12-14)^2 + (18-14)^2 + (14-14)^2 + (14-14)^2 + (16-14)^2 + (17-14)^2 + (11-14)^2 + (12-14)^2 + (15-14)^2 + (12-14)^2 + (12-14)^2 )} \) = \sqrt{6.27} \)
Step 4 ：\( \frac{s}{\sqrt{n}} = \frac{\sqrt{6.27}}{\sqrt{16}} = 0.6175 \)
Step 5 ：\( t^{*} = 1.753 \) (based on 90\% confidence level and \( n-1 = 15 \) degrees of freedom)
Step 6 ：\( \text{CI }= (\overline{x} - t^* \frac{s}{\sqrt{n}}, \overline{x} + t^* \frac{s}{\sqrt{n}}) = (14 - 1.753(0.6175), 14 + 1.753(0.6175)) = (13.08, 14.92) \)
